How many grams of aluminum sulfate Would be to med 5050g 550, completely reacted with aluminum?
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To solve this problem, we first need to determine the molar ratio between aluminum sulfate (Al2(SO4)3) and aluminum (Al) in the reaction.
From the balanced chemical equation:
2Al2(SO4)3 + 6Al -> 4Al2(SO4)3
This means that 2 moles of aluminum sulfate react with 6 moles of aluminum.
Next, we need to calculate the molar mass of aluminum sulfate:
(2x27) + (3x32) + (12x16) = 342 g/mol
Now we can calculate the number of moles of aluminum sulfate in 5050 g:
5050 g / 342 g/mol = 14.78 moles
Using the mole ratio from the balanced chemical equation, we can determine the moles of aluminum produced:
14.78 moles Al2(SO4)3 x (6 moles Al / 2 moles Al2(SO4)3) = 44.34 moles Al
Finally, we can convert moles of aluminum to grams:
44.34 moles Al x 27 g/mol = 1197.18 g
Therefore, 1197.18 grams of aluminum would be needed to react completely with 5050 grams of aluminum sulfate.